To calculate the points that Bill has after several rounds, we can use the equation he has, which is:
\[ y = 10 + 10x \]
Now let's fill in the table for Bill's points after rounds 2, 3, and 5:
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For \( x = 2 \): \[ y = 10 + 10(2) = 10 + 20 = 30 \]
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For \( x = 3 \): \[ y = 10 + 10(3) = 10 + 30 = 40 \]
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For \( x = 5 \): \[ y = 10 + 10(5) = 10 + 50 = 60 \]
Now we can summarize the points Bill has:
| Rounds (x) | Points (y) | |------------|------------| | 2 | 30 | | 3 | 40 | | 5 | 60 |
So, the filled table for Bill is:
- For 2 rounds: 30 points
- For 3 rounds: 40 points
- For 5 rounds: 60 points
Now, to determine if there is a time when Bill and Hector will have the same number of points, we can set their equations equal to each other:
Hector's points: \[ y = 100 - 5x \]
Bill's points: \[ y = 10 + 10x \]
Setting them equal: \[ 100 - 5x = 10 + 10x \]
Now, solve for \( x \): \[ 100 - 10 = 10x + 5x \] \[ 90 = 15x \] \[ x = \frac{90}{15} = 6 \]
So they will have the same number of points after 6 rounds.
At \( x = 6 \): Calculating Hector's points: \[ y = 100 - 5(6) = 100 - 30 = 70 \]
Calculating Bill's points: \[ y = 10 + 10(6) = 10 + 60 = 70 \]
Therefore, both Bill and Hector will have 70 points after 6 rounds.